There are 3 taps, A, B and C, in a tank. These can fill the tank in 10 h, 20 h and 25 h, respectively. At first, all three taps are opened simultaneously. After 2 h, tap C is closed and tap A and B keep running. After 4 h, tap B is also closed. The remaining tank is filled by tap A alone. Find the percentage of work done by tap A itself.
a) 36
b) 72
c) 108
d) 216
Sol:
Calculations: The capacity of the tank
= LCM(10, 20 and 25) = 100
Efficiency of A = 100/10 = 10
Efficiency of B = 100/20 = 5
Efficiency of C = 100/25 = 4
Tank filled in 2 hours by A, B, and C = (10 + 5 + 4) × 2 = 38 unit
According to the question,
After 4 hours from the beginning, tap B is also closed Tank filled in 2 hours by A & B = (10 + 5) × 2 = 30 unit Now, tap B is also closed Remiang capacity of tank = 100 – 38 – 30 = 32 unit These 32 unit is filled by A alone. So, the total work done by A = Work done in (1st 2 hr + next 2 hr + 32 unit) = 10 × 2 + 10 × 2 + 32 = 72 unit Percentage of a tank filled by A = (72/100) × 100 = 72%. Hence, The Required value is 72%.
